4
$\begingroup$

Predict the major product of the given reaction

I have no idea what's happening as in the product (the answer is (a)) there is a new ring being formed on the left benzene. But as I studied after $\ce{H2O}$ leaves, a carbocation is then formed on stabilized by the right ring.

Can someone please explain the rest of the steps of this reaction? The thing I don't get is how does the three-membered ring opening happens. I attempted the question, but the answer that I came up with isn't the one matching in the book. Where did I go wrong with the mechanism?

attempted mechanism

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Here is a hint. Form the most stable cation, migrate a C-C bond and do a Nazarov cyclization (Google it). To the rest of the audience, allow aditya to work on a solution before posting an answer. $\endgroup$
    – user55119
    Dec 10, 2019 at 21:53
  • $\begingroup$ Thanks I tried to attempt the question as per your suggestion but I didn't came to the right conclusion can you have a look on the mechanism as I have added it in the question. $\endgroup$
    – aditya prakash
    Dec 11, 2019 at 3:49

1 Answer 1

Reset to default
10
$\begingroup$

Your first and third structures are good but the arrow pushing in the second structure doesn't get the job done. As you have noted, diol 1 forms carbocation 2a the more stable of the two options. Now the lefthand cyclopropane bond with its pair of electrons migrates to the cationic center to form resonance-stabilized species 2b and on to ketone 3. This type of structure is a candidate for a Nazarov cyclization. The arrows in 4 may be drawn in the other direction as well. Rearomatization of 5 by loss of a proton and tautomerization leads to the product 6. [Note: Structures 2b and 4 are identical. Just a reprotonation of 3.]

ADDENDUM: To answer your original question, yes it is a pinacol rearrangement. The bond that migrates (2a --> 2b) just happens to be attached via the cyclopropane ring to the site to which it is migrating. So, 1 --> 3 is a pinacol rearrangement.

enter image description here

Improve this answer
$\endgroup$
5
  • $\begingroup$ Ok I understood most of it but I don't really get the breaking of cyclopropane and it's bond migration as I dont think I have ever studied the bond shifting or migration if a carb-cation is formed on a 3 membered cyclic ring can u please suggest some study material or something like that for it $\endgroup$
    – aditya prakash
    Dec 11, 2019 at 8:04
  • $\begingroup$ All I ever read was if there is a carb-cation on a carbon atom attached to it it undergoes dancing resonance and is highly stable but never read something about bond breaking if carb-cation is form on it $\endgroup$
    – aditya prakash
    Dec 11, 2019 at 8:07
  • $\begingroup$ aditya: I split structure 2 into 2a and 2b.. One could have migrated the lefthand cyclopropane bond to give the tertiary carbocation where the gem-dimethyl group is located. This species also leads to 3. Structure 2b is the better choice IMO. $\endgroup$
    – user55119
    Dec 11, 2019 at 14:50
  • $\begingroup$ wouldn't the carbocation be more stable near An? Could you please explain the step which leads to 2b? $\endgroup$
    – ibuprofen
    Dec 14, 2019 at 3:15
  • $\begingroup$ Yes, but now the cyclopropane ring has a chance to collapse and remove strain. The An cation is more stable than the Ph cation but that doesn't mean that An can't react. For 2b move the lefthand cyclopropane bond to the right--follow the red arrow-- and now you have a resonance stabilized double bond. 2b either deprotonates to form 3 or, since 2b = 4, undergoes Nazarov cyclization. $\endgroup$
    – user55119
    Dec 14, 2019 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.